Stochastic matrix - semisimple eigenvalue

Question

How can we prove that eigenvalue $\lambda=1$ is semisimple for any arbitrary stochastic matrix? $$\lambda \:\text{semisimple}\iff (algebraic\, multiplicity \,\lambda = geometric \,multiplicity\, \lambda)$$

Answer 1

EDIT. This old proof proves nothing. Sorry for the OP.

Let $A$ be a stochastic matrix. There exists a permutation matrix $P$ s.t. $PAP^{-1} = \left( \begin{smallmatrix} B_1 & * & * & \cdots & * \\ 0 & B_2 & * & \cdots & * \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \cdots & * \\ 0 & 0 & 0 & \cdots & B_h \end{smallmatrix} \right)$ where the $B_i$'s are either irreducible or zero. $spectrum(A)=\{sp(B_1),\cdots,{s}p(B_h)\}$. If $1\in{s}p(B_i)$ then $B_i$ is irreducible and therefore $1$ is a SIMPLE eigenvalue of $B_i$.

Unfortunately, we can only conclude that the number of indices $i$ s.t. $1\in sp(B_i)$ is equal to the algebraic multiplicity of $1$.

A correct proof is as follows.

$\textbf{Proposition.}$ If $A$ is stochastic, then all the eigenvalues of $A$ with modulus $1$ are semi simple.

$\textbf{Proof.}$ For every $k$, $A^k$ is stochastic; then $||A^k||_{\infty}=1$ and the sequence $(A^k)_k$ is bounded. Then each Jordan block of $A$ associated to an eigenvalue of modulus $1$ is $1\times 1$. $\square$